Answer:
C
Explanation: you have to observe to have an educated guess about something you test it
Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.80 g Al is mixed with 0.23 g Cl2. (a) What is the limiting reactant? Cl2 Al (b) What is the maximum amount of AlCl3, in grams, that can be produced? g AlCl3
Answer: a) [tex]Cl_2[/tex] is the limiting reagent
b) 0.27 g of [tex]AlCl_3[/tex] will be produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]
[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
According to stoichiometry :
3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]Al[/tex]
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]AlCl_3[/tex]
Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]
Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.
Solid potassium chromate is slowly added to 175 mL of a lead(II) nitrate solution until the concentration of chromate ion is 0.0366 M. The maximum amount of lead ion remaining in solution is
Answer:
[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.
Explanation:
K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)
The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.
Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.
Assuming all PbCrO₄ formed in the 175 ml solution remained ionized, then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.
To test if saturation occurs, Qsp must be > than Ksp ...
=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 => This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation. That is ...
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³
C(i) --- 0.00M 0.0336M
ΔC --- +x +x
C(eq) --- x 0.0336 + x ≅ 0.0336M
Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³
∴[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.
Answer:
[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.
Explanation:
K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)
The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.
Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.
By assumption if we say, all PbCrO₄ formed in the 175 ml solution remained ionized,
Then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.
To test if saturation occurs,
Qsp must be > than Ksp ...
=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 =>
This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation.
i.e .........
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³
C(i) --- 0.00M 0.0336M
ΔC --- +x +x
C(eq) --- x
0.0336 + x ≅ 0.0336M
Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³
Therefore;
[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.
True or False
During Solar Maximums, the thermosphere is bigger because it is warmer.
A lab group was supposed to make mL of a acid solution by mixing a solution, a solution, and a solution. However, the solution was mislabeled, and was actually a solution, so the lab group ended up with mL of a acid solution, instead. What are the volumes of the solutions that should have been mixed?
Question: The question is not complete. Find below the complete question and the answer.
Alab group was supposed to make 14 mL of a 36% acid solution by mixing a 20% solution, a 26% solution, and a 42% solution. However, the 20% solution was mislabeled, and was actually a 10% solution, so the lab group ended up with 14 mL of a 34% acid solution, instead. If the augmented matrix that represents the system of equations is given below, what are the volumes of the solutions that should have been mixed? mL
Volume of 20% solution= ?
Volume of 26% solution = ?
Volume of 42% solution= ? Round to the nearest whole number ml
Answer:
Volume of 20% solution= 3 mL
Volume of 26% solution = 1 mL
Volume of 42% solution= 10 mL
Explanation:
Find attached of the calculations.
Be sure to answer all parts. The concentration of Cu2 ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na2S) solution to 0.700 L of the water. The molecular equation is
This is an incomplete question, here is a complete question.
The concentration of Cu²⁺ ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na₂S)solution to 0.700 L of the water.
The molecular equation is:
[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]
Write the net ionic equation and calculate the molar concentration of Cu²⁺ in the water sample if 0.0177 g of solid CuS is formed.
Answer :
The net ionic equation will be,
[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]
The concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The molecular equation is:
[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]2Na^+(aq)+S^{2-}(aq)+Cu^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CuS(s)+2Na^+(aq)+SO_4^{2-}(aq)[/tex]
In this equation, [tex]Na^+\text{ and }SO_4^{2-}[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]
Now we have to calculate the mass of [tex]CuSO_4[/tex]
Molar mass of [tex]CuSO_4[/tex] = 159.5 g/mol
Molar mass of CuS is = 95.5 g/mol
From the balanced chemical reaction we conclude that,
As, 95.5 g of CuS produces from 159.5 g [tex]CuSO_4[/tex]
As, 0.0177 g of CuS produces from [tex]\frac{159.5}{95.5}\times 0.0177=0.0296g[/tex] [tex]CuSO_4[/tex]
Now we have to calculate the concentration of [tex]CuSO_4[/tex]
[tex]\text{Concentration}=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4\times \text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Concentration}=\frac{0.0296g}{159.5g/mol\times 0.700L}=2.65\times 10^{-4}M[/tex]
Concentration of [tex]Cu^{2+}[/tex] = [tex]2.65\times 10^{-4}M[/tex]
Therefore, the concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]
What is the freezing point of a solution of ethylene glycol, a nonelectrolyte, that contains 59.0 g of (CH2OH)2 dissolved in 543 g of water? Use molar masses with at least as many significant figures as the data given.
Answer:
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
Explanation:
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]
where,
[tex]T[/tex] = Freezing point of solvent
[tex]T_f[/tex] = Freezing point of solution
[tex]\Delta T_f[/tex] =depression in freezing point
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
Mass of ethylene glycol = 59.0 g
Molar mass of ethylene glycol = 62.1 g/mol
Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)
[tex]K_f[/tex] =1.86°C/m ,
[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]
[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]
[tex]\Delta T_f=3.26^oC[/tex]
Freezing point of pure water = T = 0°C
Freezing point of solution = [tex]T_f[/tex]
[tex]\Delta T_f=T-T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb calorimeter is 420.J∘C and the heat of combustion at constant volume of the sample is −3374kJmol, calculate the final temperature of the reaction in Celsius. The specific heat capacity of water is 4.184 Jg ∘C.
Answer:
22.7
Explanation:
First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:
ΔHC=qrxnn
We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:
qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J
The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:
qrxn=−(qwater+qbomb)
The heat absorbed by the water can be calculated using the specific heat of water:
qwater=mcΔT
The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:
qbomb=CΔT
Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:
−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]
Distribute the terms of each multiplication and simplify:
−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]
Add the like terms and simplify:
−7916 J=−2930.4 J∘C×Tfinal+58608 J
Finally, solve for Tfinal:
−66524 J=−2930.4 J∘C×Tfinal
Tfinal=22.701∘C
The answer should have three significant figures, so round to 22.7∘C.
A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600g of water. From the given information, the final temperature of the reaction is 22.71° C
From the given parameters:
mass of the sample C7H5N2O6 = 0.500 gmass of water = 600 ginitial temperature = 20.0° Cheat capacity of bomb calorimeter [tex]\mathbf{c_{bc}}[/tex] = 420 J/° Cheat of combustion [tex]\mathbf{\Delta H^0_c}[/tex] = -3374 kJ/molspecific heat capacity of water c = 4.184 J/g° CTO find the final temperature = ???From the listed parameters, the first step is to determine the number of moles of the sample by using the relation:
[tex]\mathbf{number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]
[tex]\mathbf{number \ of \ moles = \dfrac{0.500 \ g}{213 \ g/mol}}[/tex]
number of moles of C7H5N2O6 = 0.00235 moles
However, the heat of combustion is the amount or quantity of heat energy released when one mole of a sample is burned.
Mathematically;
[tex]\mathbf{\Delta H_c =\dfrac{q_{rxn} }{ n}}[/tex]
where;
[tex]\mathbf{ q_{rxn}}[/tex] = quantity of heat released
n = number of moles
Making [tex]\mathbf{ q_{rxn}}[/tex] the subject of the formula:
[tex]\mathbf{ q_{rxn}= \Delta H_{rxn} \times n }[/tex]
[tex]\mathbf{ q_{rxn}=-3374 \ kJ/moles \times 0.00235 \ moles }[/tex]
[tex]\mathbf{ q_{rxn}=-7.9289 \ kJ }[/tex]
[tex]\mathbf{ q_{rxn}=-7928.9 \ J }[/tex]
The heat released [tex]\mathbf{ q_{rxn}}[/tex] = heat absorbed by bomb calorimeter + heat absorbed by water
∴
[tex]\mathbf{ q_{rxn}= -(m\times c_{bc} \Delta T + c_{water} \times \Delta T)}[/tex]
[tex]\mathbf{-7928.9 \ J = -( 420 J/^0C \times \Delta T + 600 \ g \times 4.18 \ J/g ^0 C\times \Delta T)}[/tex]
[tex]\mathbf{7928.9 = (2928 \Delta T )}[/tex]
[tex]\mathbf{ \Delta T = \dfrac{7928.9}{2928} }[/tex]
[tex]\mathbf{ \Delta T =2 .71 ^0 \ C}[/tex]
Since ΔT = [tex]\mathbf{T_2 - T_1}[/tex]
2.71° C = T₂ - 20° C
T₂ = 20° C + 2.71° C
T₂ = 22.71° C
Therefore, we can conclude that the final temperature of the reaction is:
22.71° C
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The solubility of aspirin in water is 1 g per 300 mL at 25 degrees celsius. Assuming that your crystallization and washing with water were done at this temperature, what weight of aspirin did you lose in the filtrate and washings? How much was your percent yield lowered by this loss?
Answer:
The weight of aspirin lost is [tex]W =0.03267(V_A +V_B) N[/tex]
The percent yield is lowered by [tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]
Explanation:
From the question we are told that
Since the solubility of aspirin in water is 1 g per 300 mL it implies that after crystallization the solution would contain 1 g for every 300mL of water at 25°
Let assume that the volume of the solution is [tex]V_A[/tex]
The the aspirin lost after filtration would be
[tex]= \frac{1}{300} * V_A[/tex]
[tex]=\frac{V_A}{300}g[/tex]
Let assume that you used water of volume [tex]V_B[/tex] to wash the crystallized aspirin then the lost during washing would be
[tex]= \frac{1}{300} * V_B[/tex]
[tex]= \frac{V_B}{300}[/tex]
So the total loss is
[tex]= \frac{V_A}{300} + \frac{V_B}{300}[/tex]
[tex]\frac{V_A +V_B}{300}g[/tex]
So the weight of aspirin lost denoted by W is
[tex]W = \frac{V_A +V_B}{300} *9.8[/tex]
[tex]W =0.03267(V_A +V_B) N[/tex]
Let denote How much was your percent yield lowered by this loss by A
So
[tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]
Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting material and the other 2 are very different. What could these 2 nonstarting material spots be? Use structures and words. Hint: think about the nucleophilic addition of the hydride.
Answer:
Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)
Explanation:
Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.
In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.
Answer:
Explanation:
find the solution below
Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)
Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
The balanced redox reaction is
5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)
The unbalanced redox reaction is :
AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
The above equation can be balanced by ensuring the atom of the elements
on the left hand side is equal to those on the right hand side.
We have 3 atoms of Mn on the left side of the equation and 1 atom on the
right hand side.This will be balanced by putting 3 in front of MnO₄− as shown
below:
AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
We have 3 atoms of O on the left hand side and 12 atoms of O on the right
hand side. This is balanced by putting 5 in front of AsO₂− and 2 in front of
H₂O as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
We have 4 atoms of H on the left hand side and 1 atom of H on the right
hand side.We can balance by putting 4 in front of H⁺ as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)
We have 5 atoms of As on the left hand side and 1 atom of As on the right
hand side. We can balance by putting 5 in front of As as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)
The equation is therefore now balanced as the number of atoms of the
element on the left hand side are equal with those on the right hand side.
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A chemist places 10.0 grams of dry ice (CO 2 ) in a 2.5 L vacuum at 25˚C.
After it all sublimates, what would be the pressure inside the container?
Answer:
P = 2.25 atm
Explanation:
Given data:
Mass of dry ice = 10 g
Volume of container = 2.5 L
Temperature = 25°C (25+273=298 K)
Pressure inside container = ?
Solution:
First of all we will calculate the number of moles of dry ice.
Number of moles = mass / molar mass
Number of moles = 10 g/ 44 g/mol
Number of moles = 0.23 mol
Now we will determine the pressure.
PV = nRT
P = nRT/V
P = 0.23 mol × 0.0821 atm.L/mol.K × 298 K / 2.5 L
P = 5.63 atm / 2.5
P = 2.25 atm
A 0.881 g sample of a diprotic acid is dissolved in water and titrated with 0.160 M NaOH . What is the molar mass of the acid if 35.8 mL of the NaOH solution is required to neutralize the sample
Final answer:
To find the molar mass of the diprotic acid, divide the mass of the sample by the number of moles. The molar mass of the acid is 307.03 g/mol.
Explanation:
To find the molar mass of the diprotic acid, we can use the information from the titration. First, we need to determine the number of moles of NaOH used in the titration. The volume of NaOH solution used is 35.8 mL, which is equal to 0.0358 L. The molarity of the NaOH solution is 0.160 M.
Therefore, the number of moles of NaOH used is 0.0358 L x 0.160 mol/L = 0.005728 mol.
Since the diprotic acid is diprotic, it can donate two moles of H+ ions per mole of acid. Therefore, the number of moles of the diprotic acid is half of the number of moles of NaOH used, which is 0.005728 mol / 2 = 0.002864 mol.
To calculate the molar mass of the acid, we need to divide the mass of the sample by the number of moles. The mass of the acid sample is 0.881 g. Therefore, the molar mass of the acid is 0.881 g / 0.002864 mol = 307.03 g/mol.
Because of the radioactive decay of uranium and thorium in rocks and soil, radium-228, a decay product of Thorium-232, can be found in drinking water. This isotope has a half-life of 5.75 years and an atomic number of 88. If Ra-228 undergoes beta decay, what would the atomic number of the new element be? What would the mass number of this isotope be? Explain your reasoning (e.g. Explain what happens during beta decay).
Answer: The atomic number and mass number of the new element formed is 89 and 228 respectively.
Explanation:
Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron. The released beta particle is also known as electron.
In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the beta decay of Ra-228 follows:
[tex]_{88}^{228}\tyextrm{Ra}\rightarrow _{89}^{228}\textrm{Ac}+_{-1}^0\beta[/tex]
Hence, the atomic number and mass number of the new element formed is 89 and 228 respectively.
Identify the option below that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction. Select the correct answer below: When exposed to air, sodium reacts completely overnight, whereas iron barely reacts. Five 4-inch diameter saplings burn more rapidly than a 20-inch tree of the same height. Food spoils more quickly on a kitchen counter than in a refrigerator. A Bunsen burner is used in the laboratory to increase the speed of reactions.
The intrinsic chemical properties of a reactant can affect the rate of reaction, as seen in the example of food spoiling more quickly on a kitchen counter than in a refrigerator due to temperature differences.
Explanation:The option that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction is Food spoils more quickly on a kitchen counter than in a refrigerator. This is because the temperature difference between the kitchen counter and the refrigerator affects the rate of chemical reactions. Chemical reactions typically occur faster at higher temperatures, so food spoils more quickly when exposed to the higher temperature of the kitchen counter compared to the lower temperature inside the refrigerator.
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A 19.13 gram sample of chromium is heated in the presence of excess bromine. A metal bromide is formed with a mass of 77.92 g. Determine the empirical formula of the metal bromide.
Answer:
The empirical formula of the compound is [tex]CrBr_2[/tex].
Explanation:
Mass of chromium = 19.13 g
Mass metal bromide formed = 77.92 g
Mass of bromine in metal bromide = 77.92 g - 19.13 g = 58.79 g
Moles of chromium metal :
[tex]=\frac{19.13 g}{52 g/mol}=0.3679 mol[/tex]
Moles of bromine:
[tex]=\frac{58.79 g}{80 g/mol}=0.7349 mol[/tex]
For empirical formula divide the smallest number of moles of element from the all the moles of elements:
chromium : [tex]\frac{0.3679 mol}{0.3679}=1[/tex]
bromine : [tex]\frac{0.7349 mol}{0.3679}=2[/tex]
The empirical formula of the compound is [tex]CrBr_2[/tex].
To find the empirical formula, the moles of each element involved must be determined. When the sample weights are converted to moles, the ratio of chromium (Cr) to bromine (Br) in the bromide compound is found to be 1:2, making the empirical formula CrBr2.
To determine the empirical formula of the compound, we first need to identify the moles of each element. The chromium sample weighs 19.13g. Using the atomic weight of chromium, which is 52.00g/mol, the number of moles of chromium is 19.13g / 52.00g/mol = 0.368 mol.
Considering the entire mass of the metal bromide, which is 77.92 g, the mass of bromine in the compound must be 77.92 g - 19.13 g = 58.79 g. As the atomic weight of bromine is approximately 79.90g/mol, we can find that the moles of bromine are 58.79 g / 79.90 g/mol = 0.736 mol.
The ratio of Cr to Br in the compound is 0.368 : 0.736, which is approximately 1 : 2 when we divide both numbers by the smallest to obtain the simplest whole number ratio. Therefore, the empirical formula of the compound is CrBr2.
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PLEASE HELP I HAVE A TIME LIMIT
A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. What was the pH of the solution that the chemist prepared, to the nearest thousandth?
Answer:
13.279
Explanation:
24g/56.1056g/mol= 0.42 (mol KOH)
0.42 mol KOH/ 2.25 L= 0.19 M KOH
-log(0.19)= pOH= 0.72
pH=14-pOH
pH=14-0.72=13.279
sorry it's not the best explanation but since you're on a time crunch i hope this helps
A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. The pH of the solution that the chemist prepared, to the nearest thousandth is 13.279.
pH is a numerical indicator of how acidic or basic aqueous or some other liquid solutions are. The phrase, which is frequently used during chemistry, biology, or agronomy, converts hydrogen ion concentrations.
The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per liter, making it neutral.
24g/56.1056g/mol= 0.42 (mol KOH)
0.42 mol KOH/ 2.25 L= 0.19 M KOH
-log(0.19)= pOH= 0.72
pH=14-pOH
pH=14-0.72
=13.279
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What is the pH of a solution with hydroxide ion concentration of 0.005
Answer:
pH = 11.7
Explanation:
pOH= -log [OH]=-log[0.05]
=2.3
pOH+pH= 14
pH= 14-2.3= 11.7
At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rate constant of 0.0100 s − 1 . 0.0100 s−1. The reverse reaction is first order in B B and the rate constant is 0.0610 s − 1 . 0.0610 s−1. What is the value of the equilibrium constant for the reaction A − ⇀ ↽ − B A↽−−⇀B at this temperature
Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
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A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 150 kJ of work on the ideal gas. It is observed that the temperature of the ideal gas remains constant during this process as a result of heat transfer between the system and the surroundings at 30°C. Determine the entropy change of the ideal gas.
Answer:
Entropy change is zero
Explanation:
∆S=S2-S1=0
We're S is entropy
A gaseous hydrocarbon (a compound that contains only hydrogen and carbon) is found to be 11 % hydrogen by mass. a. Find the empirical formula for the compound.A gaseous hydrocarbon (a compound that contains only hydrogen and carbon) is found to be 11 % hydrogen by mass. a. Find the empirical formula for the compound.
Final answer:
To find the empirical formula of the gaseous hydrocarbon, assume 100 grams of the compound. Since it is 11% hydrogen by mass, it contains 11 grams of hydrogen. The remaining mass is carbon. Divide the moles of each element by the smallest number of moles to get the empirical formula CH₂. The molecular formula cannot be determined without the molar mass.
Explanation:
Empirical Formula Calculation
To find the empirical formula, assume 100 grams of the gaseous hydrocarbon. Since it is 11% hydrogen by mass, this means it contains 11 grams of hydrogen. The remaining mass, 89 grams, must be carbon. The molar mass of hydrogen is 1 g/mol, and the molar mass of carbon is 12 g/mol. Divide the moles of each element by the smallest number of moles to get the empirical formula. In this case, 11 g H / 1 g/mol = 11 mol H and 89 g C / 12 g/mol = 7.42 mol C. Dividing both by the smallest number of moles, we get the empirical formula CH₂.
To determine the molecular formula, you'll need the molar mass of the compound. However, this information is not provided in the question. Without the molar mass, it is not possible to determine the molecular formula.
An aqueous solution has [C6H5COOH] = 0.110 M and [Ca(C6H5COO)2] = 0.200 M. Ka = 6.3 × 10-5 for C6H5COOH. The solution volume is 5.00 L. What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added? Group of answer choices 4.81 4.86 4.75 4.70 4.65
Answer:
The answer is 4.8659
Explanation:
Given:
[C₆H₅COOH]=0.11M
[C₆H₅COO]=2*0.2=0.4M
Ka=6.3x10⁻⁵
First, calculate the pKa:
[tex]pKa=-logKa=-log(6.3x10^{-5} )=4.2007[/tex]
The pH is:
[tex]pH=pKa+log\frac{C6H5COO]}{[C6H5COOH]} =4.2007+log\frac{0.4}{0.11} =4.7614[/tex]
Like the volume is 5L, the volume of C₆H₅COO is x, then, the volume of C₆H₅COOH is 5-x
[tex]4.7614=4.2007+log\frac{0.4x}{0.11*(5-x)}[/tex]
[tex]0.5607=log\frac{0.4x}{0.11*(5-x)}[/tex]
Solving for x:
x=2.49L=2490mL of C₆H₅COO
2510mL of C₆H₅COOH
The milimoles of C₆H₅COOH and C₆H₅COO is:
nC₆H₅COOH=(0.11*2510)-50=226.1mmol
nC₆H₅COO=(0.4*2490)+50=1046mmol
The pH is:
[tex]pH=4.2007+log\frac{1046}{226.1} =4.8659[/tex]
Answer:
4.86
Explanation:
The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.
Please kindly check attachment for the step by step solution of the given problem.
A particular type of fundamental particle decays by transforming into an electron e- and a positron e . Suppose the decaying particle is at rest in a uniform magnetic field B of magnitude 3.64 mT and the e- and e move away from the decay point in the paths lying in a plane perpendicular to B. How long after the decay do the e- and e collide
Answer:
Check the explanation
Explanation:
Lorenz force exerted on electron by magnetic field is
F=e*v*H, where e=1.602E-19 C, H=3.64E-3 T, v is speed of electron;
? meanwhile Lorenz force is centripetal force F=m*v^2/r, where mass of electron m=9.11E-31 kg, r is radius of the path of electron;
? therefore F=F; e*v*H = m*v^2/r; eH=m*(v/r), hence
v/r = eH/m =w is angular speed of electron, hence
T=2pi/w =2pi*m/(eH) is period of rotation of electron;
e- and e+ starting at the same point are moving in the same circular path and in opposite directions, and should meet in T/2 time;
T/2 = pi*m/(eH) = pi*9.11E-31 /(1.602E-19 *3.48E-3) =5.13E-9 s;
In all neutral atoms, there are equal numbers of ___________________________. Group of answer choices electrons and protons neutrons and electrons protons and neutrons allotropes and electrons
In a neutral atom, there are equal numbers of electrons and protons, which gives the atom its neutral charge. Neutrons do not factor into the atom's charge, and allotropes are different forms of an element, not part of its atomic structure.
Explanation:In a neutral atom, there are equal numbers of electrons and protons. The number of protons in an atom's nucleus determines the atomic number, and the atom's charge is neutral because the positive charge of the protons is balanced by the negative charge of the electrons. Neutrons, on the other hand, do not carry an electronic charge and are not necessarily equal in number to the protons or electrons. Concerning allotropes, they are different forms of the same element and it's not directly related to this question about atomic structure.
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In neutral atoms, there are equal numbers of electrons and protons. Each element, when neutral, has a characteristic number of electrons that matches its atomic number (the number of protons). Neutrons are electrically neutral and their number does not affect the atom's charge.
Explanation:In all neutral atoms, there are equal numbers of electrons and protons. This is because the number of protons, which carry a positive charge, and the number of electrons, which carry a negative charge, balance each other in a neutral atom, resulting in an overall charge of zero.
Each element, when electrically neutral, has a characteristic number of electrons equal to its atomic number, which is the number of protons it contains. For example, in a carbon-12 atom (which is neutral), there are six protons and six electrons, yielding a net charge of zero.
It's important to note that while neutrons also exist within an atom, they are electrically neutral. Therefore, the number of neutrons does not affect the atom's charge and doesn't necessarily match the number of protons or electrons.
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What is the hydrogen-ion concentration of the ph is 3.7
Answer:
the answer 37
Explanation:
Calculate the amount of heat needed to melt of solid octane () and bring it to a temperature of . Round your answer to significant digits. Also, be sure your answer contains a unit symbol.
Answer:
85.0 kJ
Explanation:
Calculate the amount of heat needed to melt 160. g of solid octane (C8H18 ) and bring it to a temperature of 99.2 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
To melt 160 g of octane and bring it's temperature to 99.2°C
(from literature)
Heat of fusion of Octane = 20.740 kJ/mol
Melting point of octane = -57°C
Boiling point of Octane = 125.6 °C
Molar mass of octane = 114.23 g/mol
Heat capacity of octane = 255.68 J/K.mol
So, it is evident that Octane is still a liquid at 99.2°C.
So, the required heat is the heat required to melt octane and raise the temperature of Octane liquid to 99.2°C
First, we convert the mass of octane given to number of moles as the heat parameters provided by literature are given in molar units.
Number of moles = (mass)/(Molar mass)
Number of moles of octane = (160/114.23) = 1.401 moles
Heat required to melt the octane = nL = (1.401×20.740) = 29.05674 kJ
Heat required to raise the temperature of already melted octane from its melting temperature of -57°C to 99.2°C
= nCΔT
n = 1.401 moles
C = 255.68 J/K.mol
ΔT = (99.2 - (-57)) = 156.2°C (same as a temperature difference of 156.2 K)
Heat required to raise the temperature of already melted octane from -57°C to 99.2°C
= (1.401×255.68×156.2)
= 55,952.04 J = 55.952 kJ
Total heat required to melt the 160 g of Octane and raise its temperature to 99.2°C
= 29.05674 + 55.952 = 85.01 kJ = 85.0 kJ
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The heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process - melting and then heating. You use the latent heat of fusion and specific heat capacity of octane to calculate the total heat required.
Explanation:
The amount of heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process – melting, and heating. The process should be looked at in separate steps, similar to how the enthalpy of combustion is used to calculate the heat produced in the combustion of 1.00 L of isooctane in the reference provided. Note that different substances have different heat capacities and heat of fusion, so the associated values (likely given in the problem or textbook) for octane specifically would need to be used to get the final answer.
First, you would need to know the latent heat of fusion, which is the amount of energy required to change one kilogram of the substance from solid to liquid without changing its temperature. Then, you would multiply the mass of the octane by the latent heat of fusion to find out how much energy is required for it to go from solid to liquid.
Once the octane is in liquid form, we then need to heat it to the desired temperature. To do this, you use the specific heat capacity formula – the amount of energy required to raise one kilogram of the substance by one degree Celsius. Multiply the mass of the material by the specific heat capacity of the octane and the change in temperature (final temperature – initial temperature).
The added heat required for both of these processes would give you your final answer.
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Step 3: Prepare Seven Solutions to Establish a pH
Scale
0.1 M
NaOH
The acids and bases shown right cover a range of pH values. Use
what you know about acids, bases, and concentration to label the
test tubes, in order, from most acidic to most basic.
0.1 M
HCI
Distilled
Water
0.00001 M
NaOH
0.001 M
NaOH
0.001 M
HCI
0.00001 M
HCI
Answer: starting on the left:
1. 0.1M HCI
2. 0.001M HCI
3. 0.00001M HCI
4. distilled water
5. 0.00001 NaOH
6. 0.001M NaOH
7. 0.1M NaOH
Explanation:
The solutions arranged from most acidic to most basic would be: 0.1 M HCl, 0.001 M HCl, 0.00001 M HCl, Distilled Water, 0.00001 M NaOH, 0.001 M NaOH, and finally 0.1 M NaOH.
Explanation:The pH of a solution depends on the concentration of hydronium ions (H+) and hydroxide ions (OH-). In this case, a lower concentration means less acidity or basicity. The solutions arranged in order from most acidic to most basic would be:
0.1 M HCl - strongest acid because it has the highest concentration of H+ ions. 0.001 M HCl - slightly less acidic than 0.1 M HCl. 0.00001 M HCl - least acidic as the H+ ion concentration is lowest amongst the acids. Distilled Water - neutral with a pH of 7 because it has an equal concentration of H+ and OH- ions. 0.00001 M NaOH - least basic as the OH- ion concentration is the smallest amongst the bases. 0.001 M NaOH - more basic than 0.00001 M NaOH because it has a higher concentration of OH- ions. 0.1 M NaOH - most basic solution as it has the highest concentration of OH- ions. Learn more about pH Scale here:
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1. A sample of hydrogen gas is collected over water in a flask at 23.5 C. The pressure in the flask is equilibrated with the atmospheric pressure. The atmospheric (barometric) pressure is reported as 746.7 torr. Determine the pressure of hydrogen gas in the flask.
Answer:
the pressure of hydrogen gas in the flask is 725 torr
Explanation:
the solution is in the attached Word file
The pressure of hydrogen gas in the flask is [tex]\( 725.6 \, \text{torr} \)[/tex].
To determine the pressure of hydrogen gas collected over water, we need to account for the vapor pressure of water at the given temperature. The total pressure in the flask is the sum of the partial pressures of hydrogen gas and water vapor.
The steps to solve this are:
1. Find the vapor pressure of water at the given temperature (23.5°C):
The vapor pressure of water at 23.5°C is approximately 21.1 torr (this value can be found in tables of water vapor pressures).
2. Calculate the pressure of hydrogen gas:
The total pressure in the flask is the atmospheric pressure, which is 746.7 torr. The pressure of hydrogen gas is the total pressure minus the vapor pressure of water.
[tex]\[\text{Pressure of hydrogen gas} = \text{Total pressure} - \text{Vapor pressure of water}\][/tex]
[tex]\[P_{\text{H}_2} = 746.7 \, \text{torr} - 21.1 \, \text{torr}\][/tex]
[tex]\[P_{\text{H}_2} = 725.6 \, \text{torr}\][/tex]
ANSWER ASAP
Suppose that 10.00 HCl of unknown concentration is neutralized by 20.00 mL of a 1.50 M NaOH solution. Determine the concentration of the HCl solution?
a. 0.0750 M HCl
b. 0.150 M HCl
c. 3.00 M HCl
d. 1.50 M HCl
Answer:
c. 3.00 M HCl
Explanation:
From dilution formula
C1V1 = C2V2
C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml
Substitute and Simplify
C1×10= 1.5×20
C1= 3.00M
Assume that Aluminum and Silver Sulfide are the starting substances (reactants) in the reaction: a. Write a balanced chemical equation describing the "re-creation" of silver, using the information in the case study. b. State the names of the products that are produced from this reaction. c. What type of reaction(s) is/are being represented by the chemical reaction you wrote in part (a)? d. Is the reaction in part (a) an oxidation-reduction (redox) reaction? e. If this is a redox reaction, then identify the following: What is undergoing oxidation (what is being oxidized)? Support your answer using oxidation numbers. What is undergoing reduction (what is being reduced)? Support your answer using oxidation numbers. What is the reducing agent? What is the oxidizing agent?
Answer:
Check Explanation
Explanation:
a) The balanced chemical reaction between Aluminium and Silver Sulfide is represented below
2Al + 3Ag₂S → Al₂S₃ + 6Ag
Aluminium displaces Silver from Silver sulfide because it is higher than Silver in the electrochemical series.
b) Names of the products from this reaction
Ag is called Silver metal. Free Silver metal.
Al₂S₃ is called Aluminium Sulfide.
c) This reaction is a single-displacement reaction because an element directly displaces and replaces another element in a compound.
It is also a redox reaction (reduction-oxidation reaction) because there are species being oxidized and reduced simultaneously!
d) Yes, this reaction is an oxidation-reduction (redox) reaction because there are species being oxidized and reduced simultaneously!
e) The specie that is being oxidized is said to undergo oxidation. And oxidation is defined as the loss of electrons, thereby leading to an increase in oxidation number.
In this reaction, it is evident that Aluminium undergoes oxidation as its oxidation number increases from 0 in the free state to +3 when it displaces Silver and becomes Aluminium sulfide.
Al → Al³⁺
0 → +3 (Oxidation)
The specie that is being reduced is said to undergo reduction. And reduction is defined as the gain of electrons, thereby leading to a decrease in oxidation number.
In this reaction, it is evident that the silver ion undergoes reduction as its oxidation number decreases from +1 in the Silver Sulfide compound to 0 when it is displaced and becomes Silver in free state.
Ag⁺ → Ag
+1 → 0 (Reduction)
The reducing agent is the specie that brings about reduction. Since Silver ion in Silver sulfide is reduced, Aluminium is the reducing agent that initiates the reduction process.
The oxidation agent is the specie that brings about oxidation. Since, Aluminium is the specie that undergoes oxidation, the oxidizing agent is the Silver Sulfide that brings about the oxidation.
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Liquefied natural gas (LNG) is transported in very large tankers, stored as liquid in equilibrium with its vapor at approximately atmospheric pressure. If LNG is essentially pure methane, the storage temperature then is about 111.4 K, the normal boiling point of methane. The enormous amount of cold liquid can in principle serve as a heat sink for an onboard heat engine. Energy discarded to the LNG serves for its vaporization. If the heat source is ambient air at 300 K, and if the efficiency of a heat engine is 61% of its Carnot value, estimate the vaporization rate in moles vaporized per kJ of power output. For methane,
Answer:
0.2 mol/kJ
Explanation:
Methane is stored at a temperature of 111.4K, [tex]T_c[/tex]. The heat source to vapourization of methane is ambient air which is at 300 K. [tex]T_H[/tex]
Estimate the vaporization rate at the efficiency of heat engine 60% of its carnot value.
Calculate the vaporization rate from the given data by relation shown below:
[tex]Vaporization rate = \frac{Q_c}{[\frac{\delta H_n^{lv}}{W}]} ..........(1)[/tex]
here,
[tex]Q_c[/tex] is the heat at temperature [tex]T_c, \delta H_n^{lv}[/tex] is the phase transition enthalpy of methane and W is the work
Calculate [tex]Q_c[/tex] from the equation shown below:
[tex]Q_c = Q_n (1 - \eta_{HE}) .............(2)[/tex]
where Q_n is the heat at temperature of [tex]T_n[/tex] and [tex]\eta_{HE}[/tex] is the efficiency of heat engine
calculate [tex]Q_H[/tex] from the relation shown below:
[tex]Q_H = \frac{W}{\eta_{HE}} ..........(3)[/tex]
calculate the heat engine efficiency from the given carnot engine efficiency as shown below
[tex]\eta_{HE} = 0.6 \times \eta_{carnot} ............(4)[/tex]
here, [tex]\eta_{carnot}[/tex] is the carnot engine efficiency
[tex]\eta{carnot} = 1 - \frac{T_c}{T_H}[/tex]
substituting the values of temperature, we have
[tex]= 1 - \frac{111.4K}{300K}\\= 0.629\\[/tex]
substitute values of [tex]\eta_{carnot}[/tex] in equation 4, we get
[tex]\eta_{HE} =0.6 \times \eta_{carnot}\\ = 0.6 \times 0.629\\ = 0.3774\\\\[/tex]
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